show in detailed steps limit x--->6 sq.r(x +(sq.r.x +(sq.r.x +....to infinty)?

y = sqr(x +sqr(x +sqr(x +....to infinty)

Square both sides of this forumla:

y^2 = x + sqr(x +sqr(x +sqr(x +....to infinty)

Since x goes to infinity, one less step is still infinity. Therefore:

y^2 = x + y

Since the limit is as x approaches 6, substitute x=6

y^2 = 6 + y

Let's move this into quadratic form:

y^2 - y - 6 = 0

Using the Quadratic Equation with a=1, b=-1, c=-6

x = (-b +/- sqr( b^2 -4ac )) / 2a

x = ( 1 +/- sqr(1 - 4*1*(-6))) / 2

x = ( 1 +/- (sqr(25)) / 2

x = ( 1 +/- 5 ) / 2 = 3 or -2

Since a square root, by definition, returns a non-negative number, the limit is 3.

23

Let y=sq.r(x +(sq.r.x +(sq.r.x +....to infinty)

square both sides

y^2=x+(sq.r.x +(sq.r.x +....to infinty)

y^2=x+y

Lim x->6

y^2=6+y

y^2-y-6=0

y^2-3y+2y-6=0

y(y-3)+2(y-3)=0

(y-3)(y+2)=0

y-3=0 --> y=3

y+2=0--> y=-2

Therefore,

limit x--->6 sq.r(x +(sq.r.x +(sq.r.x +....to infinty)

=3, -2